Complex numbers

(a) Because of the absolute value we may assume that \(r > 0\) by replacing \(r\) with \(-r\) if necessary. Now use AM-GM inequality or the fact that \((\sqrt{r}-\sqrt{1 / r})^{2} \geq 0 .\) since \(x+x^{-1}=\frac{\sqrt{5}+1}{2}<2\) given \(x\) must be a non-real complex number.

(b) Let \(\theta=\frac{\pi}{5} .\) Then \(\sin (2 \theta)=\sin (\pi-2 \theta)=\sin (3 \theta) .\) Using the formulas for \(\sin (2 \theta)\) and \(\sin (3 \theta),\) canceling \(\sin \theta\left(\text { it is nonzero) and substituting } \sin^{2} \theta=1-\cos ^{2} \theta,\right.\) gives the quadratic equation \(4 \cos ^{2} \theta-2 \cos \theta-1=0 .\) since \(\cos \theta > 0\), we get \(\cos \theta=\frac{\sqrt{5}+1}{4}\)

(c) Let \(x=d e^{i \alpha}=d(\cos \alpha+i \sin \alpha) .\) Then \(x^{-1}=d^{-1} e^{-i \alpha}=d^{-1}(\cos \alpha-i \sin \alpha) .\) Adding and using that \(x+x^{-1}=\frac{\sqrt{5}+1}{2}=2 \cos \left(\frac{\pi}{5}\right),\) we get \(d=1\) and \(\alpha=\pm \theta .\) So \(x=e^{\pm \frac{1 \pi}{5}}\).
\begin{align} x^{2012}+x^{-2012} &= 2 \cos \left(\frac{2012 \pi}{5}\right)\\ &=2 \cos \left(402 \pi+\frac{2 \pi}{5}\right)\\ &=2 \cos \left(\frac{2 \pi}{5}\right)\\ &=2 \cos ^{2}\left(\frac{\pi}{5}\right)-1 \\ &=\frac{\sqrt{5}-1}{2}\\ \end{align}

Since \(z^n=1\), we have \(z^n-1=0\).

\[ z^n-1 = (z^{n-1} + z^{n-2} + \cdots + 1)(z-1) = 0 \]

For \(n\geq 2 \), \(z\neq 1\). So the first factor must be zero. This proves the statement.

\begin{align} A &:= \cos 1^\circ + \cos 41^\circ + \cos 81^\circ + \cdots + \cos 321^\circ \\ B &:= \sin 1^\circ + \sin 41^\circ + \sin 81^\circ + \cdots + \sin 321^\circ \\ \end{align}

Notice that \(40^\circ=2\pi/9\). Let \( \theta = 1^\circ = \pi/180 \). Then:

\[ A+iB = e^{i\theta} \left( \sum_{k=0}^{8} e^{ \frac{2\pi i}{9}k } \right) \]

From problem 2(i), we know that RHS of the above equation is zero. Since \(A\) and \(B\) are real numbers, both of them must be individually zero. In particular, \(A=0\), which proves the statement.

\begin{align} \left(b^{2}\right)^{2}+\left(\frac{i^{2}}{b^{2}}\right)^{2}&=6\\ \left(b^{2}+\frac{i^{2}}{b^{2}}\right)^{2}&=4\\ b^{2}+\frac{i^{2}}{b^{2}}&=\pm 2 \end{align}

Let us know look at the quantity we want to compute:

\begin{align} \left(b+\frac{i}{b}\right)^{16}&=\left(b^{2}+\frac{i^{2}}{b^{2}}+2 i\right)^{8}\\ &=(\pm 2+2i)^{8}\\ &=2^{8}\left(\sqrt{2} e^{\frac{i \pi}{4}}\right)^{8} \text{ or } 2^{8}\left(\sqrt{2} e^{\frac{3i \pi}{4}}\right)^{8} \\ &=2^{12}\\ &=4096 \end{align}

a) \(e^{i A}+e^{i B}+e^{i C}=0\) by taking the vector sum of the three points on the unit circle.

b) \(\cos A+\cos B+\cos C=0=\) real part of \(e^{i A}+e^{i B}+e^{i C},\) which is 0 by part a.

c) \(\cos 2 A+\cos 2 B+\cos 2 C=0\) because the points \(e^{2 i A}, e^{2 i B}, e^{2 i C}\) on the unit circle also form an equilateral triangle in the complex plane, since taking \(B=A+(2 \pi / 3), C=A+(4 \pi / 3)\), we get \(2 B=2 A+(4 \pi / 3)\) and \(2 C=2 A+(8 \pi / 3)=2 A+(2 \pi / 3)+2 \pi\). The last term \(2 \pi\) does not change the position of the point.

d) \(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=\frac{3}{2}\) because, using the formula for \(\cos 2 \theta\) in part \(c,\) we get \(\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=\sin ^{2} A+\sin ^{2} B+\sin ^{2} C\) and the sum of the LHS and the RHS in this equation is \(3 .\)

(i) Each \(e^{i\theta}\) can take a maximum value of 1, which is attained when \(\theta=0\). Hence, the maximum average is also 1.

(ii) To get the minimum, place the 13 points on the vertices of a regular cyclic polygon. The average corresponds to the center of the polygon which is \((0,0)\). Hence, the minimum value \(|A|\) can take is 0.

(i) ||\(z_{1}|-| z_{2}||<\left|z_{1}+z_{2}\right| .\) One way: using triangle inequality for \(z_{1}+z_{2}\) and \(-z_{2}\) we get \(\left|z_{1}\right| \leq\left|z_{1}+z_{2}\right|+\left|-z_{2}\right|\) and so \(\left|z_{1}\right|-\left|z_{2}\right| \leq\left|z_{1}+z_{2}\right| .\) Now we may take absolute value on the LHS because switching \(z_{1}\) and \(z_{2}\) keeps RHS the same. For equality, \(z_{1}+z_{2}\) and \(-z_{2}\) must point in the same direction, so \(z_{1}\) and \(z_{2}\) must be along the same line. But they are in quadrants 2 and \(3,\) so this cannot happen.

(ii) \(z_{2}\) must be obtained by rotating \(z_{1}\) by angle \(\pi / 3,\) say in the counterclockwise direction (otherwise interchange the two). Then \(z_{2}=z_{1} e^{\frac{\pi i}{3}}\). Then \(z_{1}^{2}+z_{2}^{2}=z_{1}^{2}\left(1+e^{\frac{2 \pi i}{3}}\right)\) and \(z_{1} z_{2}=z_{1}^{2} e^{\frac{\pi i}{3}} . \quad\) Now \(1+e^{\frac{2 \pi i}{3}}=e^{\frac{\pi i}{3}}\) (see by calculation or picture), so we have in fact \(z_{1}^{2}+z_{2}^{2}=z_{1} z_{2}\)

(iii) We have \(\prod_{i=1, \ldots, 7}\left(x-z_{i}\right)=\frac{x^{8}-1}{x-1}=1+x+\ldots+x^{7}\). Putting \(x=1\) gives \(\prod_{1=1, \ldots, 7}\left(1-z_{i}\right)=8\)

(a) In general, the equation \(z^n = re^{i\theta}\) has \(n\) solutions given by: \[ r^{1 / n} \exp \left[\frac{i(\theta+2 k \pi)}{n}\right] \text{ for each } 0 \leq k \leq n-1 \] The given equation has 2018 complex solutions, since we can express the complex number in the RHS as \(re^{i\theta}\) for some small \(\theta\).

(b) Instead of looking at the given equation, first consider the solutions to: \(x^{2018}=r\). Two of them are real values: \( r^{1/2018} \) and \( -(r)^{1/2018} \). The other 2016 are distributed equally in the four quadrants, 504 each.

If we rotate the solutions to \(x^{2018}=r\) by a tiny angle in the counter-clockwise direction, we get the solutions to the given equation. (The value of \(r\) being \( \sqrt{ 2018^{2018\cdot 2} + 1^2 } \)). This gives 505 values each in the first and third quadrant but still 504 in the second and fourth quadrant.

Such roots can be expressed as follows

\[ w=\frac{\cos (2 \pi k)}{2019}+i \frac{\sin (2 \pi k)}{2019} \quad \text { for } k=0,\pm 1, \ldots,\pm 1009 \]

Therefore,

\[ |w+1|^{2}=2+2 \frac{\cos (2 \pi k)}{2019} \]

Hence we want to find all \(k\) such that

\[ \frac{\cos (2 \pi k)}{2019} \geq \frac{1}{\sqrt{2}} \]

Which is same as

\[ \begin{array}{l} \quad\left|\frac{2 \pi k}{2019}\right| \leq \frac{\pi}{4} \\ \text { i.e. }|k| \leq 252 \end{array} \]

So there are 505 solutions.